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Difference between revisions of "1986 AJHSME Problems/Problem 25"

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==Solution==
 
==Solution==
  
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There seems to be no better way to solve this other than just find each of those, so that's what we do.
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===Solution 1===
  
 
From <math>1</math> to <math>101</math> there are <math>\left\lfloor \frac{101}{2} \right\rfloor = 50</math> (see [[Floor function|floor function]]) multiples of <math>2</math>, and their average is  
 
From <math>1</math> to <math>101</math> there are <math>\left\lfloor \frac{101}{2} \right\rfloor = 50</math> (see [[Floor function|floor function]]) multiples of <math>2</math>, and their average is  
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Similarly, we can find that the average of the multiples of <math>3</math> between <math>1</math> and <math>101</math> is <math>51</math>, the average of the multiples of <math>4</math> is <math>52</math>, the average of the multiples of <math>5</math> is <math>52.5</math>, and the average of the multiples of <math>6</math> is <math>51</math>, so the one with the largest average is <math>\boxed{\text{D}}</math>
 
Similarly, we can find that the average of the multiples of <math>3</math> between <math>1</math> and <math>101</math> is <math>51</math>, the average of the multiples of <math>4</math> is <math>52</math>, the average of the multiples of <math>5</math> is <math>52.5</math>, and the average of the multiples of <math>6</math> is <math>51</math>, so the one with the largest average is <math>\boxed{\text{D}}</math>
  
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''Note: You can just add the first term and the last term, and see which one is the biggest. 2+100=102, 3+99=102, 4+100=104, 5+100=105, 6+96=102. Therefore, the answer is multiples of five because it has the highest number.''
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===Solution 2===
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You can just add the first term and the last term, and see which one is the biggest. 2+100=102, 3+99=102, 4+100=104, 5+100=105, 6+96=102. Therefore, the answer is multiples of five because it has the highest number.
  
 
==See Also==
 
==See Also==

Revision as of 15:44, 23 May 2010

Problem

Which of the following sets of whole numbers has the largest average?

$\text{(A)}\ \text{multiples of 2 between 1 and 101} \qquad \text{(B)}\ \text{multiples of 3 between 1 and 101}$

$\text{(C)}\ \text{multiples of 4 between 1 and 101} \qquad \text{(D)}\ \text{multiples of 5 between 1 and 101}$

$\text{(E)}\ \text{multiples of 6 between 1 and 101}$

Solution

Solution 1

From $1$ to $101$ there are $\left\lfloor \frac{101}{2} \right\rfloor = 50$ (see floor function) multiples of $2$, and their average is \begin{align*} \frac{2\cdot 1+2\cdot 2+2\cdot 3+\cdots + 2\cdot 50}{50} &= \frac{2(1+2+3+\cdots +50)}{50} \\ &= \frac{2\cdot \frac{50\cdot 51}{2}}{50} \\ &= \frac{2\cdot 51}{2} \\ &= 51 \\ \end{align*}

Similarly, we can find that the average of the multiples of $3$ between $1$ and $101$ is $51$, the average of the multiples of $4$ is $52$, the average of the multiples of $5$ is $52.5$, and the average of the multiples of $6$ is $51$, so the one with the largest average is $\boxed{\text{D}}$

Solution 2

You can just add the first term and the last term, and see which one is the biggest. 2+100=102, 3+99=102, 4+100=104, 5+100=105, 6+96=102. Therefore, the answer is multiples of five because it has the highest number.

See Also

1986 AJHSME (Problems • Answer Key • Resources)
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